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This approach uses a combination of Depth First Search (DFS) and memoization to solve the problem efficiently. We explore each cell, attempting to find the longest increasing path starting from that cell. To avoid recalculating the path length for each call from the same cell, we use memoization to store already computed results.
The time complexity is O(m * n) because each cell is computed once and cached. The space complexity is also O(m * n) due to the memoization table.
1def longestIncreasingPath(matrix):
2 if not matrix or not matrix[0]:
3 return 0
4
5 m, n = len(matrix), len(matrix[0])
6 memo = [[-1] * n for _ in range(m)]
7
8 def dfs(x, y):
9 if memo[x][y] != -1:
10 return memo[x][y]
11
12 directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
13 max_len = 1
14
15 for dx, dy in directions:
16 nx, ny = x + dx, y + dy
17 if 0 <= nx < m and 0 <= ny < n and matrix[nx][ny] > matrix[x][y]:
18 max_len = max(max_len, 1 + dfs(nx, ny))
19
20 memo[x][y] = max_len
21 return max_len
22
23 return max(dfs(x, y) for x in range(m) for y in range(n))In this Python solution, we define a helper function dfs that performs DFS on the matrix starting from a given cell. We utilize a memo table to store the longest path from each cell to avoid redundant calculations, improving performance significantly. The main function starts by iterating over each cell to compute the maximum length path by considering each cell as a starting point.
This approach systematically calculates the longest increasing path dynamically by first iterating over the matrix and then updating results based on previously computed values. By using a dynamic programming table, we determine, for each cell, the longest chain it can contribute to incrementally.
The time complexity is O(m * n) due to the fact that each matrix cell can be a starting point and is only calculated once with memoization guidance. The space complexity is O(m * n) because a result is stored for each cell in the DP table.
1public class Solution {
2 public int LongestIncreasingPath(int[][] matrix) {
3 if (matrix == null || matrix.Length == 0) return 0;
4 int m = matrix.Length, n = matrix[0].Length;
int[][] dp = new int[m][];
for (int i = 0; i < m; i++) dp[i] = new int[n];
int maxLen = 0;
int dfs(int x, int y) {
if (dp[x][y] != 0) return dp[x][y];
int[][] directions = new int[][]{new int[]{-1, 0}, new int[]{1, 0}, new int[]{0, -1}, new int[]{0, 1}};
dp[x][y] = 1;
foreach (var dir in directions) {
int nx = x + dir[0], ny = y + dir[1];
if (nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[nx][ny] > matrix[x][y]) {
dp[x][y] = Math.Max(dp[x][y], 1 + dfs(nx, ny));
}
}
return dp[x][y];
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
maxLen = Math.Max(maxLen, dfs(i, j));
}
}
return maxLen;
}
}This C# solution follows a similar logic by using dynamic programming principles. The DFS function explores potential paths from each cell, leveraging an array dp to store already calculated results and determine the longest increasing path starting from a specific point.